6.3 Maxwell–Mohr Method 169
a
R R
R
A B
C
+t 1 +t 2
DB
h 0
b
x
X=1
x R
y j
R
y ds
A B
HA
C
+t 1 +t 2
DB
dj
Fig. 6.14 Curvilinear bar. Design diagram and unit state
Unit loadXD 1 (Fig.6.14b) corresponds to required horizontal displacement at
B. In the unit state reactionHAD 1 and internal forces due to unit loadXare as
follows
NNkD 1 sin'I MNkD 1 yD 1 Rsin': (b)
Thus equation (a) for displacement atBbecomes
BD ̨
t 1 Ct 2
2
ZR
0
. 1 sin'/dsC ̨
jt 1 t 2 j
h 0
ZR
0
.Rsin'/ds: (c)
Integration is performed along a curvilinear road of lengthR. In the polar
coordinates
d'D
ds
R
; sin'D
y
R
;
the limits of integration become 0 –
BD ̨
t 1 Ct 2
2
Z
0
. 1 sin'/Rd'C ̨
jt 1 t 2 j
h 0
Z
0
.Rsin'/Rd': (d)
Thus, for required displacement we get the following expression
BD ̨
t 1 Ct 2
2
.2R/C ̨
jt 1 t 2 j
h 0
.2R^2 /D ̨R .t 1 Ct 2 /2 ̨R^2
jt 1 t 2 j
h 0
:(e)
Negative sign in (e) means that unit forceXproduces negative work on the real
displacement, i.e., the displacement of the pointBdue to temperature changes is
directed from left to right.
For uniform change of temperature (i.e., when gradients for indoor and outdoor
temperatures are the same), i.e.,t 1 Dt 2 , a differencet 1 – t 2 D 0 and only first term of
(6.15) or (e) should be taken into account. In this case, the horizontal displacement
equals toBD2 ̨Rt.